Problem: Find $\lim_{\theta\to 0}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{2}$ (Choice C) C $\dfrac{1}{4}$ (Choice D) D The limit doesn't exist
Solution: Substituting $\theta=0$ into $\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\sin^2(\theta)$ in our expression, let's rewrite it using the Pythagorean identity, $\sin^2(\theta)+\cos^2(\theta)=1$ : $\begin{aligned} &\phantom{=}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)} \\\\ &=\dfrac{1-\cos(\theta)}{2(1-\cos^2(\theta))} \gray{\text{The Pythagorean identity}} \\\\ &=\dfrac{1-\cos(\theta)}{2(1+\cos(\theta))(1-\cos(\theta))} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{\cancel{1-\cos(\theta)}}{2(1+\cos(\theta))\cancel{(1-\cos(\theta))}} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{1}{2(1+\cos(\theta))}\text{, for }\theta\neq \{...,-4\pi, -2\pi, 0, 2\pi, 4\pi,...\} \end{aligned}$ This means that the two expressions have the same value for all $\theta $ -values (in their domains) except for $2k\pi$ for any integer $k$, and specifically $0$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}=\dfrac{1}{2(1+\cos(\theta))}$ for all $\theta$ -values in the interval $(-1,1)$ except for $\theta=0$. Therefore, $\lim_{\theta\to 0}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}=\lim_{\theta\to 0}\dfrac{1}{2(1+\cos(\theta))}=\dfrac{1}{4}$. (The last limit was found using direct substitution.) In conclusion, $\lim_{\theta\to 0}\dfrac{1-\cos(\theta)}{2\sin^2(\theta)}=\dfrac{1}{4}$.